Answer to Question #345488 in Statistics and Probability for Shy

Question

A group of students got the following scores in a test: 6, 9, 12, 15, 18, and 21. Consider samples of size 3 that can be drawn from this population.

A. List all the possible samples and the corresponding mean.

B. Construct the sampling distribution of the sample means.

Answer 1

We have population values 6,9,12,15,18,21, population size N=6 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{6+9+12+15+18+21}{6}=13.5$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{6}(56.25+20.25+2.25$$ $$+2.25+20.25+56.25)=26.25$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{26.25}\approx5.1235$$

A. Select a random sample of size 3 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{6}C_3=20.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 6,9,12 & 9 \\ \hdashline 2 & 6,9,15 & 10 \\ \hdashline 3 & 6,9,18 & 11 \\ \hdashline 4 & 6,9,21 & 12 \\ \hdashline 5 & 6,12,15 & 11 \\ \hdashline 6 & 6,12,18 & 12 \\ \hdashline 7 & 6,12, 21 & 13 \\ \hdashline 8 & 6,15,18 & 13 \\ \hdashline 9 & 6,15,21 & 14 \\ \hdashline 10 & 6, 18,21 & 15 \\ \hdashline 11 & 9,12,15 & 12 \\ \hdashline 12 & 9, 12,18 & 13 \\ \hdashline 13 & 9, 12, 21 & 14 \\ \hdashline 14 & 9,15,18 & 14 \\ \hdashline 15 & 9,15,21 & 15 \\ \hdashline 16 & 9, 18,21 & 16 \\ \hdashline 17 & 12, 15,18 & 15 \\ \hdashline 18 & 12, 15,21 & 16 \\ \hdashline 19 & 12, 18,21 & 17 \\ \hdashline 20 & 15, 18,21 & 18 \\ \hdashline \end{array}$$

B.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 9 & 1/20 & 9/20 & 81/20 \\ \hdashline 10 & 1/20 & 10/20 & 100/20 \\ \hdashline 11 & 2/20 & 22/20 & 242/20 \\ \hdashline 12 & 3/20 & 36/20 & 432/20 \\ \hdashline 13 & 3/20 & 39/20 & 507/20 \\ \hdashline 14 & 3/20 & 42/20 & 588/20 \\ \hdashline 15 & 3/20 & 45/20 & 675/20 \\ \hdashline 16 & 2/20 & 32/20 & 512/20 \\ \hdashline 17 & 1/20 & 17/20 & 289/20 \\ \hdashline 18 & 1/20 & 18/20 & 324/20 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{270}{20}=13.5=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{3750}{20}-(13.5)^2=5.25= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$\sigma_{\bar{X}}=\sqrt{5.25}\approx2.2913$$